Find the volume of the solid in the first octant bounded by the sphere \rho = 22, the coordinate...
Question:
Find the volume of the solid in the first octant bounded by the sphere {eq}\rho = 22 {/eq}, the coordinate planes, and the cones {eq}\phi = \pi / 6 {/eq} and {eq}\phi = \pi / 3 {/eq}
Upper V equals
Finding the Volume:
The objective is to find the volume of the solid lies in the first octant over the region {eq}E {/eq}
The volume formula is,
{eq}V = \iiint_{E} f \left( \rho, \phi, \theta \right) \ dV {/eq}
By spherical coordinates,
{eq}dV = \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta {/eq}
The given cone range lies between {eq}\frac{\pi}{6} \ to \ \frac{\pi}{3} {/eq} and {eq}\rho = 22 {/eq}.
Given that, the region lies in the first octant. so that, {eq}\theta {/eq} range varoius from {eq}0 \ to \ \frac{\pi}{2} {/eq}.
Answer and Explanation: 1
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View this answerThe sphere is {eq}\rho = 22 {/eq}
The cone is:
{eq}\phi = \frac{\pi}{6} \\ \phi = \frac{\pi}{3} {/eq}
The solid lies in the first octant. So...
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Volumes of Shapes: Definition & Examples
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Chapter 11 / Lesson 9
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In this lesson, learn the definition of volume and how to find the volume of objects of various shapes. Learn from various solved volume examples.
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